Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How do you know if there isn't a solution to a calculus optimization problem?

  1. Jan 11, 2009 #1
    My teacher was saying that it is possible to have no solution to an optimization problem, and I was curious about how this could be possible. Could someone please explain and possibly give an example?
  2. jcsd
  3. Jan 11, 2009 #2


    User Avatar
    Science Advisor
    Gold Member

    Here are two examples:

    Find maximum value of function [itex]f(x) = x [/itex]

    Note that nowhere is the derivative zero or undefined (no critical points). Note also that no boundary is given. At best you can say the maximum occurs in the limit as [itex]x \to \infty[/itex].

    Find maximum value of function [itex]f(x)=x^3[/itex]

    Here we have a critical point at [itex] x = 0[/itex] but note it is an inflection point. Testing points to each side will tell you it is not a local min or max. Again given no boundaries you get no absolute maximum or minimum when you check the behavior as [itex] x \to \pm \infty[/itex]

    Third example: Find the absolute maxima and minima of [itex] f(x) = 1/x [/itex] on the interval [itex] -1 \le x \le 1[/itex]. Again you see a critical point at x=0 where the derivative is undefined. However the function is also undefined there and you see that it has a vertical asymptote there. Since the asymptote is two sided going to [itex]\infty[/itex] on one side and [itex] -\infty[/itex] on the other then there is no maximum or minimum value.
  4. Jan 11, 2009 #3
    Well my teachr gave us the example:
    A Piece of paper with an area of 100in is to be rolled into an open cylinder with a maximum volume. What dimensions should the paper have to accomplish this. When we solved it in class he said there was no solution but I don't understand why. Could you explain?
  5. Jan 11, 2009 #4
    In general, this would be because the function that describes the volume of the cylinder is proportional to r2. The derivative of this function, which tells you where the maximum would be, is proportional to r. The derivative function will only be zero when r = 0, the trivial case. Since there is no other example where the derivative function is zero, i.e. no other critical points, there cannot be a maximum to the volume.
  6. Jan 11, 2009 #5
    I'm sorry but I'm not really understanding what are you are saying here. Could you possibly rephrase it or simplify it?
  7. Jan 12, 2009 #6
    The function representing the volume is monotonic, it doesnt have any critical points.
    V=50r in this case or V=b*50/2pi where b is one of the sides of the paper. It only means that the volume will grow unbounded because you can increase b as much as you want, (the condition for the other side a=100/b can still be satisfied.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook